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Define the unit ramp function by
1. Determine the Laplace transform of H(t)
2. Use the Laplace transform to solve the ODE
1. To determine the Laplace transform, do we use
∫ (note the difference in integration limits). I know both integrals will give the same answer. But I
am confused because H(t) is also defined for t<0 and therefore it seems a bit unnatural to let the lower limit of integration be 0.
In any case, the answer to this part is
2. Taking the Laplace transform of the given equation yields
Thus to find the solution y(t) we must determine the inverse Laplace
transformation of the above. This is where I am having trouble since
rewriting the expression for Y(s) seems to give different expressions for
y(t), non of which actually satisfies (*) when substituted. A few of these
options are given here:
u(t) denotes the Heaviside function. Alternatively, we know from
part 1 that the Laplace transform of H(s)=1/s^2 so we could also have
None of these solutions actually seems to give (*) if we calculate the second
derivative of y(t) and substitute it into (*), the RHS does not turn out to be H(t‐1)
My question therefore is:
1. What is the correct Laplace transform in this case.
2. Please show step by step how to obtain this (i.e. if we have to use some
method like using residues, please show how to do this in detail, because I
don't really know how to do this.
p.s. Maple seems to give even more different Inverse Laplace transforms, but I
am not really interested in how to do it with Maple, since I want to know how to
do it by hand.
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The Laplace transform and its inverse can be very useful in solving PDEs subject to given boundary conditions. The solution comprises a 1 page word document with equations written in Mathtype illustrating the use in one particular example. There are also some preliminary comments related to theory questions asked by the student.
Please see the attached file for the complete solution.
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For your first question, the Laplace transform is (usually) defined as the integral from zero. It can also be defined as the integral from , but this is much less convenient (you lose convergence of the standard functions). Consequently, in solutions obtained by ...
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